Question

A container holds 1.00 L methane gas at 660 torr pressure when the temperature is 22.0 C how much more pressure will the gas exert if the temperature is raised to 44.6 C

Answer 1

Answer:

$P_{2} = 710.537\,torr$

Explanation:

Let suppose that gas behaves ideally, the corresponding equation of state is:

$P\cdot V = n\cdot R_{u}\cdot T$

Volume and quantity of moles remain constant due to the container, so that final pressure can be found by the following relationship:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{317.75\,K}{295.15\,K}\cdot 660\,torr$

$P_{2} = 710.537\,torr$

Answer 2

Answer:

assuming it is an ideal gas, the correct answer would be that P2: 710,537 torr

Explanation:

If this gas behaves like an ideal gas, then the appropriate equation we would use is that of noble gases:

PxV = n x r x T

where p is the pressure, v the volume, t the temperature, n number of moles and r is the constant of the noble gases which is 0.082x10exp23.

Since the variables V n and r are constant for this behavior, the only thing that varies is the pressure and the temperature, that is why there is an initial and final pressure and an initial or final temperature.

We will call the initial pressure P1, and the final pressure P2, as well as the temperature.

in this way the equation that would remain is:

P1 / T1 = P2 / T2.

What we are looking for is the final pressure, that is, P2, that is why it is cleared in the equation, remaining as the final equation:

P2 = (T2 / T1) x P1